Question

Derive an expression for integrated rate law for a zero order reaction.

Answer 1

Hint :This question gives the knowledge about the zero order reaction. In zero order reaction, the reaction in which on changing the concentration of reactant, there is no change on the rate of the reaction. The zero order reaction can be explained using Haber’s process.

Complete Step By Step Answer:
In zero order reaction, the reaction in which on changing the concentration of the reactant, there is no change on the rate of the reaction. The unit for zero order reaction is $ mol{L^{ - 1}}{s^{ - 1}} $ .
The derivation for the expression of integrated rate law for zero order reaction is as follows:
Consider a reaction for zero order,
 $ A \to P $
Where $ A $ is a reactant and $ P $ is a product.
At initial time the concentration of reactant is $ a $ and the reactant is zero. At time $ t $ the concentration of the reaction changes to $ a - x $ and the concentration of products become $ x $ . Where $ a $ is the initial amount of the reactant, $ a - x $ is the left amount of the reactant and $ x $ is the amount of reactant decomposed.
The rate of the reaction is as follows:
 $ \Rightarrow r = {k_0}{\left[ A \right]^0} = \dfrac{{ - d\left[ A \right]}}{{dt}} = \dfrac{{d\left[ P \right]}}{{dt}} $
Substitute the concentration of reactant $ A $ at time $ t $ is $ a - x $ ,
 $ \Rightarrow r = \dfrac{{ - d\left[ {a - x} \right]}}{{dt}} = {k_0} $
On simplifying, we get
 $ \Rightarrow r = \dfrac{{ - da}}{{dt}} + \dfrac{{dx}}{{dt}} = {k_0} $
Neglect $ \dfrac{{ - da}}{{dt}} $ in the above equation because there is no change in $ a $ with respect to time.
 $ \Rightarrow {k_0} = \dfrac{{dx}}{{dt}} $
On rearranging the above equation, we get
 $ \Rightarrow dx = {k_0}dt $
Now, integrate the above equation, $ dx $ within limits $ 0 $ to $ x $ and $ dt $ within limits $ 0 $ to $ t $ .
 $ \Rightarrow \int\limits_0^x {dx} = {k_0}\int\limits_0^t {dt} $
On solving the above equation, we get
 $ \Rightarrow \left[ x \right]_0^x = {k_0}\left[ t \right]_0^t $
On further simplifying, we get
 $ \Rightarrow x - 0 = {k_0}\left( {t - 0} \right) $
The expression for zero order reaction will be as follows:
 $ \Rightarrow x = {k_0}t $
Therefore, the expression for integrated rate law for a zero order reaction is $ x = {k_0}t $ .

Note :
Always remember the concept that in the zero order reaction, the reaction in which on changing the concentration of reactant, there is no change on the rate of the reaction. Haber’s process follows zero order reaction.